# Combination problem with 4 groups of values

• chiraganand
In summary, there are 36 possible combinations when grouping ten different elements into four groups with a restriction on the number of combinations within a group.
chiraganand
Summary:: 10 values are divided into 4 groups and need a combination of these with restrictions placed on group size, ordering and combinations

I have a combinations question.. i have 4 group of values A , B, C and D, with A-2 values, B-3 values , C-2 values, D-3 values.
1. From each group only one value has to be taken A and B are together and C and D are together. So 1 value from A , 1 Value from B and then different combinations of C and D (maintaining 1 value each).
2. The order of the values from the groups is interchangeable that is 1-A 1-B 1-C 1-D or 1-B 1-A 1-C 1-D or 1-B 1-A 1-D 1-C so on..
I am trying to calculate the number of combinations but I am coming up short.. according to me there should be 36 different combinations but i am unable to take the flexible ordering into account. Can someone help me with this

[Moderator's note: moved from a technical forum.]

Last edited by a moderator:
The ways to group ten all different elements into A,B,C and D count
$$_{10}C_3\ \ _7C_3\ \ _ 4C_2=25,200$$
I was not able to understand what you mean to go further. I appreciate if you show examples of your operation on the set e.g.
$$\{0,1,2,3,4,5,6,7,8,9\}$$

Thanks for the reply, so for example A={0,1}, B={2,3,4}, C={5,6}, D={7,8,9}. So one combination would be 0257 0258 0259 0267 0268 0269, and so on with 03.., 04.. 12..,13..14.. and also 2075 or 31 76 .. and so on.. i hope i am making sense now? The elements from A and B have to take the first two places (1st and 2nd or 2nd and 1st) and the elements of C and D have to take the 3rd and 4th places(can also be 4th and 3rd)

The number of ways in the order ABCD: 2*3*2*3=36
The number of ways where A and B can be swapped: 2*36=72
The number of ways where C and D can also be swapped: 2*72=144
That is all I can understand from your description of the problem.

chiraganand
Thanks.
chiraganand said:
so for example A={0,1}, B={2,3,4}, C={5,6}, D={7,8,9}.
The 36 combinations are
0257
0258
0259
0267
0268
0269
0357
0358
0359
0367
0368
0369
0457
0458
0459
0467
0468
0469
1257
1258
1259
1267
1268
1269
1357
1358
1359
1367
1368
1369
1457
1458
1459
1467
1468
1469

Last edited:
chiraganand
FactChecker said:
The number of ways in the order ABCD: 2*3*2*3=36
The number of ways where A and B can be swapped: 2*36=72
The number of ways where C and D can also be swapped: 2*72=144
That is all I can understand from your description of the problem.
Yes this was the answer which I came up with later. I was wondering if there is a combination formula for the same

anuttarasammyak said:
Thanks.

The 36 combinations are
0257
0258
0259
0267
0268
0269
0357
0358
0359
0367
0368
0369
0457
0458
0459
0467
0468
0469
1257
1258
1259
1267
1268
1269
1357
1358
1359
1367
1368
1369
1457
1458
1459
1467
1468
1469
Yes i understand the combinations but as the number of elements increases, how do i calculate the total number of combinations

After you choose a division you get 36 combinations as you understand.
With all the possible divisions you get ## _{10}C_4## cases.

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## 1. What is a combination problem with 4 groups of values?

A combination problem with 4 groups of values involves selecting a specific number of items from each of the 4 groups to create a combination. This type of problem is commonly encountered in statistics, probability, and combinatorics.

## 2. How many combinations can be formed from 4 groups of values?

The number of combinations that can be formed from 4 groups of values depends on the number of items in each group. The formula for calculating the number of combinations is nCr = n! / (r! * (n-r)!), where n is the total number of items and r is the number of items being selected. In this case, n would be the sum of the items in all 4 groups.

## 3. What is the difference between a combination and a permutation?

A combination involves selecting items without regard to their order, while a permutation involves selecting items in a specific order. In other words, a combination is a selection of items, while a permutation is an arrangement of items.

## 4. How can I solve a combination problem with 4 groups of values?

To solve a combination problem with 4 groups of values, you can use the formula nCr = n! / (r! * (n-r)!). First, determine the total number of items in all 4 groups and the number of items being selected. Then, plug these values into the formula to calculate the number of combinations. Alternatively, you can use a combination calculator or a combination formula table.

## 5. Can a combination problem with 4 groups of values have repeated items?

Yes, a combination problem with 4 groups of values can have repeated items. In this case, the formula for calculating the number of combinations would be nHr = (n+r-1)! / (r! * (n-1)!), where n is the total number of items and r is the number of items being selected. This formula is known as the combination with repetition formula.

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